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3 years ago

What is the chemical formula for an amino group?

1 answer:

4 0

<span>Formula of the 20 common amino acids. The formula of an amino acid comprises, bound to a carbon (alpha carbon): a carboxyl group -COOH. an amine group -NH2.</span>

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slamgirl [31]

I will help you with answering this question.

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3 years ago

Out of 13 students how many<br> saw themselves in the<br> horoscope?

Andru [333]

Answer: Astrology is not astronomy! ... Astronomers and other scientists know that stars many light years* away have no effect on the ... Imagine a straight line drawn from Earth through the Sun and out into space way ... The 13 constellations in the zodiac. ... on the phases of the Moon), each month got a slice of the zodiac all to itself.

Explanation: Hope this helps

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3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be

DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

$n = c \cdot V$ ,

where

$n$ is the number of moles of the solute in the solution.
$c$ is the concentration of the solution. $c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1}$ for the initial solution.
$V$ is the volume of the solution. For the initial solution, $V = 135\;\textbf{mL} = 0.135\;\textbf{L}$ for the initial solution.

$n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}$ .

What's the concentration of the diluted solution?

$\displaystyle c = \frac{n}{V}$ .

$n$ is the number of solute in the solution. Diluting the solution does not influence the value of $n$ . $n = 0.03375\;\text{mol}$ for the diluted solution.
Volume of the diluted solution: $25\;\text{mL} + 135\;\text{mL} = 160\;\textbf{mL} = 0.160\;\textbf{L}$ .

Concentration of the diluted solution:

$\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}$ .

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0

3 years ago

Read 2 more answers

Помогите срочно сделать по химии промежуточную отестацию за 8 класс

il63 [147K]

Answer:

This is site for English speakers. Этот сайт на английском, поэтому вопрос могут удалить

Explanation:

1. 2)

2. 3)

3. 4) Sr

4. 3)

5. 4)

6. 2)

7. 1)

8. 4)

9. 3)

10. 3)

11. SO3, H2SO4, Na2SO4

12.

A) оксид меди (II) 2) CuO

Б) хлорид меди(II) 4) CuCl2

В) сульфит меди (II) 3) CuSO 3

Г) гидроксид меди (II) 1) Cu(OH)2

13.

1. Fe+HCl= б) FeCl 2 +H 2

2.Fe+O2= в) Fe 3 O 4

3. Fe(OH) 3 = г)Fe 2 O 3 +H 2O

4. FeCl 2 +NaOH= а) Fe(OH) 2 +NaCl

14. 2Ca + O2 = 2CaO

CaO + H2O = Ca(OH)2

Ca(OH)2 + 2HCl = CaCl2 + 2H2O

6 0

3 years ago